TRIANGLE-FREE INTERSECTION GRAPHS OF LINE
SEGMENTS WITH LARGE CHROMATIC NUMBER
arXiv:1209.1595v5 [math.CO] 26 Dec 2014
ARKADIUSZ PAWLIK, JAKUB KOZIK, TOMASZ KRAWCZYK, MICHAŁ LASOŃ,
PIOTR MICEK, WILLIAM T. TROTTER, AND BARTOSZ WALCZAK
Abstract. In the 1970s, Erdős asked whether the chromatic number of intersection graphs of line segments in the plane is bounded by a function of
their clique number. We show the answer is no. Specifically, for each positive
integer k, we construct a triangle-free family of line segments in the plane
with chromatic number greater than k. Our construction disproves a conjecture of Scott that graphs excluding induced subdivisions of any fixed graph
have chromatic number bounded by a function of their clique number.
1. Introduction
A proper coloring of a graph is an assignment of colors to the vertices of the
graph such that no two adjacent ones are assigned the same color. The minimum
number of colors sufficient to color a graph G properly is called the chromatic
number of G and denoted by χ(G). The maximum size of a clique (a set of
pairwise adjacent vertices) in a graph G is called the clique number of G and
denoted by ω(G). It is clear that χ(G) > ω(G). A class of graphs is χ-bounded
if there is a function f : N → N such that χ(G) 6 f (ω(G)) holds for any graph
G from the class1. A triangle is a clique of size 3. A graph is triangle-free if it
does not contain any triangle.
There are various constructions of triangle-free graphs with arbitrarily large
chromatic number. The first one was given by Zykov [17], and the one perhaps
best known is due to Mycielski [11]. On the other hand, in the widely studied class of perfect graphs, which include interval graphs, split graphs, chordal
graphs, comparability graphs, etc., the chromatic number and the clique number
are equal. In particular, these graphs are 2-colorable when triangle-free.
The intersection graph of a family of sets F is the graph with vertex set F
and edge set consisting of pairs of intersecting elements of F. For simplicity, we
identify the family F with its intersection graph.
The study of the relation between χ and ω for geometric intersection graphs
was initiated by Asplund and Grünbaum [1]. They proved that the families of
axis-aligned rectangles in the plane are χ-bounded. On the other hand, Burling
A journal version of this paper appeared in J. Combin. Theory Ser. B, 105:6–10, 2014.
Five authors were supported by Ministry of Science and Higher Education of Poland grant
884/N-ESF-EuroGIGA/10/2011/0 within ESF EuroGIGA project GraDR.
1This notion has been introduced by Gyárfás [5], who called the class χ-bound and the
function χ-binding. However, the term χ-bounded seems to be better established in the modern
terminology.
1
2
PAWLIK, KOZIK, KRAWCZYK, LASOŃ, MICEK, TROTTER, AND WALCZAK
[3] showed that the triangle-free families of axis-aligned boxes in R3 have arbitrarily large chromatic number.
Paul Erdős asked in the 1970s2 whether the families of line segments in the
plane are χ-bounded. Kratochvı́l and Nešetřil generalized this question to the
families of curves in the plane any two of which intersect at most once, see [7].
McGuinness [9] conjectured that the families F of bounded arcwise connected
sets in the plane such that for any S, T ∈ F, the set S ∩ T is arcwise connected
or empty are χ-bounded.
Some important special cases of Erdős’s question are known to have positive
solutions. In particular, Suk [15] proved that the families of segments intersecting a common line and the families of unit-length segments are χ-bounded,
improving results of McGuinness for triangle-free families [9].
We show that the answer to Erdős’s question is negative. Namely, for every
positive integer k, we construct a triangle-free family S of line segments in the
plane such that χ(S) > k.
A related question of Erdős3 asks whether the class of complements of intersection graphs of line segments in the plane is χ-bounded. Here the answer is
positive as shown by Pach and Törőcsik [12].
As first observed by Fox and Pach [4], our result disproves a purely graphtheoretical conjecture of Scott [14] that for every graph H, the class of graphs
excluding induced subdivisions of H is χ-bounded. To see this, take H being
the 1-subdivision of a non-planar graph, and note that no subdivision of such H
is representable as an intersection graph of segments.
2. Proof
Theorem 1. For every integer k > 1, there is a family S of line segments in
the plane with no three pairwise intersecting segments and with χ(S) > k.
We actually prove a stronger and more technical lemma, which admits a relatively compact inductive proof.
Let S be a family of line segments contained in the interior of a rectangle
R = [a, c] × [b, d]. A rectangle P = [a0 , c] × [b0 , d0 ] is a probe for (S, R) if the
following conditions are satisfied:
(i) We have a < a0 < c and b < b0 < d0 < d. Note that the right boundary of
P lies on the right boundary of R.
(ii) No line segment in S intersects the left boundary of P .
(iii) No line segment in S has an endpoint inside or on the boundary of P .
(iv) The line segments in S intersecting P are pairwise disjoint.
The reader should envision a probe as a thin rectangle entering R from the
right and intersecting an independent set of line segments. The purpose of the
restriction on the left boundary of the probe is to simplify the details of the
2An approximate date confirmed in personal communication with András Gyárfás and János
Pach; see also [5, Problem 1.9] and [2, Problem 2 in Section 9.6].
3See [5, Problem 1.10].
TRIANGLE-FREE SEGMENT GRAPHS WITH LARGE CHROMATIC NUMBER
3
Figure 1. Segments, probes and roots
construction to follow. The rectangle [a0 , c0 ] × [b0 , d0 ] with maximum c0 that is
internally disjoint from every line segment in S is the root of P .
In the argument below, we construct a family of pairwise disjoint probes for
(S, R). We illustrate such a configuration in Figure 1.
We define sequences (si )i∈N and (pi )i∈N by induction, setting s1 = p1 = 1,
si+1 = (pi + 1)si + p2i , and pi+1 = 2p2i .
Lemma 2. Let k > 1 and R be an axis-aligned rectangle with positive area.
There is a triangle-free family Sk of sk line segments in the interior of R and
a family Pk of pk pairwise disjoint probes for (Sk , R) such that for any proper
coloring φ of Sk , there is a probe P ∈ Pk for which φ uses at least k colors on
the segments in Sk intersecting P .
Proof. The proof goes by induction on k. For the base case k = 1, we pick any
non-horizontal segment inside R as the only member of S1 . The probe in P1 is
any rectangle contained in R, touching the right boundary of R and piercing the
chosen segment with its lower and upper boundaries.
Now goes the induction step: for a given rectangle R, we construct Sk+1 and
Pk+1 . First, we draw a family S = Sk inside R and let P = Pk be the associated
set of probes, as claimed by the induction hypothesis. Then, for each probe
P ∈ P, we place another copy SP of Sk with set of probes QP inside the root of
P . Finally, for every P ∈ P and every Q ∈ QP , we draw the diagonal of Q, that
is, the line segment DQ from the bottom-left corner of Q to the top-right corner
of Q. Note that DQ crosses all segments pierced by Q and no other segment.
The family Sk+1 consists of the line segments from the pk + 1 copies of Sk and
the p2k diagonals, so the total number of segments is (pk + 1)sk + p2k = sk+1 .
Now we show how the set of probes Pk+1 is constructed. For P ∈ P and
Q ∈ QP , let S(P ) be the segments in S that intersect P and SP (Q) be the
segments in SP that intersect Q. For each P ∈ P and each Q ∈ QP , we put two
probes into Pk+1 : a lower probe LQ and an upper probe UQ , both lying inside
Q extended to the right boundary of R. We choose the lower probe LQ very
4
PAWLIK, KOZIK, KRAWCZYK, LASOŃ, MICEK, TROTTER, AND WALCZAK
Figure 2. A diagonal with lower and upper probes
close to the bottom edge of Q and thin enough so that it intersects all segments
in SP (Q) but not the diagonal DQ . We choose the upper probe UQ very close
to the top edge of Q and thin enough so that it intersects the diagonal DQ but
not the segments in SP (Q). Both LQ and UQ end at the right boundary of R,
as required by the definition of probe, and thus intersect also the segments in
S(P ). Note that LQ and UQ are disjoint. By the induction hypothesis and the
placement of SP inside the root of P , the sets S(P )∪SP (Q) and S(P )∪{DQ } are
both independent, so LQ and UQ indeed satisfy the conditions for being probes.
See Figure 2 for an illustration. Clearly, the probes in Pk+1 are pairwise disjoint
and their total number is 2p2k = pk+1 .
The family Sk+1 is triangle-free, because we constructed Sk+1 by taking disjoint copies of triangle-free families and adding diagonals intersecting independent sets of segments. Let φ be a proper coloring of Sk+1 . We show that there is
a probe in Pk+1 for which φ uses at least k +1 colors on the line segments in Sk+1
intersecting that probe. Consider the restriction of φ to S, the original copy of
Sk used to launch the construction. There is a probe P ∈ P such that φ uses at
least k colors on the line segments in S intersecting P . Now, consider SP , the
copy of Sk put inside the root of P . Again, there is a probe Q ∈ QP such that
φ uses at least k colors on the segments in SP intersecting Q. If φ uses different
sets of colors on the segments intersecting P and the segments intersecting Q,
then at least k + 1 colors are used on the segments pierced by the lower probe
LQ . If φ uses the same set of colors on the segments intersecting P and those
intersecting Q, then another color must be used on the diagonal DQ , and thus φ
uses at least k + 1 colors on the segments intersecting the upper probe UQ . The smallest family S̃k of segments satisfying the conclusion of Theorem 1
that we know is obtained by taking all segments from the family Sk constructed
above and adding the diagonals of all probes in Pk . It is indeed triangle-free
as the segments in Sk intersecting every probe form an independent set. Since
every proper coloring of Sk uses at least k colors on the segments intersecting
some probe, the diagonal of this probe must receive yet another color, which
yields χ(S̃k ) > k. By an argument similar to that in the proof of Lemma 2, we
can show that the family S̃k is (k + 1)-critical, which means that χ(S̃k ) = k + 1
and removing any segment from S̃k decreases the chromatic number to k.
It should be noted that the family S̃k and the triangle-free family of axisaligned boxes in R3 with chromatic number greater than k constructed by Burling
[3] yield the same intersection graph.
TRIANGLE-FREE SEGMENT GRAPHS WITH LARGE CHROMATIC NUMBER
5
3. Remarks
An easy generalization of the presented construction shows the following: for
every arcwise connected compact set S in the plane that is not an axis-aligned
rectangle, the triangle-free families of sets obtained from S by translation and
independent scaling in two directions have unbounded chromatic number. In
particular, this gives the negative answer to a question of Gyárfás and Lehel [6]
whether the families of axis-aligned L-shapes are χ-bounded. For some sets S
(e.g. circles and square boundaries), we can even restrict the transformations
to translation and uniform scaling and still obtain graphs with arbitrarily large
chromatic number. We discuss this matter in a follow-up paper [13].
4. Problems
Problem 1. What is (asymptotically) the maximum chromatic number of a
triangle-free family of n segments in the plane?
The size of the triangle-free family S̃k of segments with χ(S̃k ) > k that we
construct in Section 2 is sk + pk . It easily follows from the inductive definition
of pk and sk that
k−1 −1
22
k−1
= p k 6 s k 6 22
− 1.
2k−1
Therefore, we have |S̃k | = Θ(2
). This shows that the maximum chromatic
number of a triangle-free family of segments of size n is of order Ω(log log n). On
the other hand, the bound of O(log n) follows from a result of McGuinness [9].
Problem 2. Is there a constant c > 0 such that every triangle-free family of n
segments in the plane contains an independent subfamily of size at least cn?
By the above-mentioned bound χ(S) = O(log n), any triangle-free family S
of segments of size n contains an independent subfamily of size Ω(n/ log n).
Update. Walczak [16] proved the negative answer to the question in Problem 2.
The chromatic number of segment intersection graphs containing no triangles and no 4-cycles is bounded, as shown by Kostochka and Nešetřil [8]. The
following problem has been proposed by Jacob Fox.
Problem 3. Do the families of segments in the plane containing no triangles
and no 5-cycles have bounded chromatic number?
Update. We have learned from Sean McGuinness that the positive answer to
the question in Problem 3 is a direct corollary to the following result of him [10,
Theorem 5.3]: if a triangle-free segment intersection graph has large chromatic
number, then it contains a vertex v with the property that the vertices at distance
2 from v induce a subgraph with large chromatic number. Indeed, any two
adjacent vertices at distance 2 from v witness a 5-cycle. Iterated application of
McGuinness’s result shows that for any ` > 5, triangle-free segment intersection
graphs with chromatic number large enough contain a cycle of length `. We are
grateful to Sean for these observations.
6
PAWLIK, KOZIK, KRAWCZYK, LASOŃ, MICEK, TROTTER, AND WALCZAK
Acknowledgments
We thank Jacob Fox and János Pach for their helpful remarks and advice.
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(Arkadiusz Pawlik, Jakub Kozik, Tomasz Krawczyk, Piotr Micek, Bartosz Walczak) Theoretical
Computer Science Department, Faculty of Mathematics and Computer Science, Jagiellonian
University, Kraków, Poland
E-mail address: {pawlik,jkozik,krawczyk,micek,walczak}@tcs.uj.edu.pl
(Michał Lasoń) Theoretical Computer Science Department, Faculty of Mathematics and Computer Science, Jagiellonian University, Kraków, Poland; Institute of Mathematics of the Polish
Academy of Sciences, Warsaw, Poland
E-mail address: [email protected]
(William T. Trotter) School of Mathematics, Georgia Institute of Technology, Atlanta, GA
30332, USA
E-mail address: [email protected]